# projectile motion

by terryaki
Tags: motion, projectile
 P: n/a I need help putting this problem into a workable equation: A projectile is launched from ground level at an angle of 12 degrees above the horizontal. It returns to ground level. To what value should the launch angle be adjusted, w/o changing the launch speed, so that the range doubles? So far I tried this: I broke the problem into its components: x & y. FOR X: {x=Vo*COS(12)t Vox=Vo*Cos(12) a=0 t=t} FOR Y: {y=? Voy=Vo*SIN(12) a=-9.8 m/s^2 t=t} Then I used t=x/Vo*COS(12), then substituted that for T in the Y parts, so that: y= tan(12)x - [(4.9 x^2)/(Vo^2*cos(12)^2)] but then I got: x=tan(12)y, which doesn't help me at all. I'm guessing I'm approaching this problem in a totally WRONG way!
 Emeritus P: 1,919 so you start out with a vector v-> and an angle theta1. x = |v->| * cos(theta1) * t y = |v->| * sin(theta1) * t - (1/2) * g * t^2 when y = 0, the projectile will be on the ground 0 = |v->| * sine(theta1) * t - (1/2) * g * t^2 t = 0 , 2 *|v->| * sin(theta1) / g when t = 0 is at the start, so the other solution is the time it hits the ground x = |v->| * cos(theta1) * 2 *|v->| * sin(theta1) / g pluging in t into the x equation and doing some math you get: x = |v->|^2 * sin( 2 * theta1) / g If you want the range to be double then it is 2 times the x equation with the new angle. x2 = 2 * |v->|^2 * sin( 2 * theta2) / g equate x and x2 to find theta2. You know theta1 = 12 degrees x = x2 , then theta2 = 5.87 degrees And there is your answer

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